Square matrix of A has nonzero determinant if and only if the characteristic equation has all nonzero roots

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Please help me with this. How do I prove that a square matrix A has non-zero determinant(nonsingular) if and only if the characteristic equation of A has all non-zero roots. Its probably very simple...

Thank you in advance

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The characteristic polynomial of $A$ is $p(\lambda)=\det (A-\lambda I)$. So, $p(0)=\det A$ and the result follows.